# Deduce PDF of a mixed distribution from the CDF

CDF given as:

$$F(x)= \begin{cases} \dfrac{1}{4}&\text{ if }& 0\leq x<\dfrac{1}{4}\\ \dfrac{1}{2}&\text{ if }&\dfrac{1}{4}\leq x<\dfrac{1}{2}\\ \dfrac{3}{4}&\text{ if }&\dfrac{1}{2}\leq x<\dfrac{3}{4}\\ \dfrac{x+3}{5}&\text{ if }&\dfrac{3}{4}\leq x<2\\ 1 &\text{ if }& x\geq2. \end{cases}$$

Then I tried to calculate PDF as given below: $$f(x)= \begin{cases} \dfrac{1}{4}&\text{ if }& 0\leq x<\dfrac{1}{4}\\ \dfrac{1}{4}&\text{ if }&\dfrac{1}{4}\leq x<\dfrac{1}{2}\\ \dfrac{1}{4}&\text{ if }&\dfrac{1}{2}\leq x<\dfrac{3}{4}\\ \dfrac{1}{5}&\text{ if }&\dfrac{3}{4}\leq x<2\\ \end{cases}$$

I am having problem in last segment of PDF. Please someone tell me. Will it be zero? And how am I suppose to find cdf as $1$ out of it?

• Something is wrong in the second line ... – Isham Jan 22 '18 at 5:01
• @Isham Yes thank you edited that out. – user459687 Jan 22 '18 at 5:03
• From the given CDF, you see the function are piecewisely constants for $x < 3/4$, with jumps of magnitude $1/4$ on $0, 1/4, 1/2$ respectively, therefore it just has 3 discrete point mass and no such pdf exist on that interval. On $(3/4, 2)$, the CDF is a linear function so it is just a continuous uniform distribution over that region. Note there are no more jumps on $3/4$ or $2$. – BGM Jan 22 '18 at 6:06
• I edited the post. Please consider the changes. – zoli Jan 22 '18 at 7:44
• Hard to answer this latter comment. The Community is expecting more care, For example, it turned out that you did not understand how to calculate the pdf. Yet, your question was not like "I was told the the pdf is the derivative of the cdf. But here, there is not derivative. Pls. help me understand what is going on with this pdf?" (With all due respect...) – zoli Jan 22 '18 at 8:15

The discrete part is $\frac14,\frac14,\frac14$ at $0$, $\frac14$, $\frac12$, respectively. The continuous part is
$$g(x)=\begin{cases}\frac15&\text{ if }&\frac34<x<2\\0&\text{ otherwise.}\end{cases}$$
• Sorry but the PDF is nowhere equal to $\frac45$. Or are you alluding to a conditional density (which would be a bad idea)? (Also, please note that "the probability that the distribution is continuous" can only be zero in the context, since the distribution is not continuous...) – Did Jan 22 '18 at 8:28